Question

find the pair of integers whose in that was once some is 10 [a-]differences is-8​

Answer 1

Answer:

Bhai -8 se +10 tak pair bana do

Answer 2

Let the two integers be x and y.

ATQ,

$x + y = 10 \: \: \: \: eqn.1$

$x - y = - 8$

$x = y - 8 \: \: \: \: eqn.2$

Put in eqn.1

$(y - 8) + y = 10$

$2y - 8 = 10$

$2y = 10 + 8 = 18$

$y = \frac{18}{2} = 9$

Put in eqn.2

$x = 9 - 8 = 1$

Be Brainly!!!!